The known function f (x) satisfies f (TaNx) = 1 Sin2x · cos2x, find the analytic formula of F (x)

The known function f (x) satisfies f (TaNx) = 1 Sin2x · cos2x, find the analytic formula of F (x)

∵f（tanx）=1
sin2x•cos2x
=4
(2sinxcosx)2
=4
sin22x
= (2
sin2x)2
=(1+tan2x
tanx)2，
∴f（x）=(1+x2
X) 2
=1
x2+x2+2（x≠0）．

If the known function f (x) satisfies f (TaNx) = 1 / sin * 2xcos * 2x, then the analytic formula of F (x) is

Let TaNx = T1 / sin * 2xcos * 2x = (SiNx ^ 2 + cosx ^ 2) * (SiNx ^ 2 + cosx ^ 2) / 2sinxcosx (cosx ^ 2-sinx ^ 2) divide by cosx ^ 2 * cosx ^ 2 to get f (TaNx) = (TaNx ^ 2 + 1) (TaNx ^ 2 + 1) / 2tanx (1-tanx ^ 2) f (T) = (T ^ 2 + 1) ^ 2 / 2T (1-T ^ 2)

It is known that the distance between two adjacent symmetry axes of an image with F (x) = sin ^ 2wx + 3coswx * cos (π / 2-wx) (W > 0) and y = f (x) is π (1) Find the value of W and monotone increasing interval of F (x) (2) In △ ABC, a, B and C are the opposite sides of angles a, B and C respectively. If a = radical 3, B = radical 2, f (a) = 3, find the angle C The process needs to be more detailed, within an hour

Cos (π / 2-wx) = sin (Wx), so f (x) = sin ^ 2wx + root 3coswx sin (Wx) so = half (root 3 + 2) times sin ^ 2wx, because the distance between two adjacent symmetry axes is π
So w = 1) find the value of W and the monotone increasing interval of F (x)
Find the angle a by F (a) = 3

It is known that the distance between two adjacent symmetry axes of an image with the function FX = 2 radical 3sin (Wx + π / 3) (W > 0, X ∈ R) is π 2 in the triangle ABC, if f (a) = 3 and BC = radical 3, find the maximum area of the triangle ABC

The distance between the two axes of symmetry is π
∴T/2=π
T=2π
w=2π/T=2π/2π=1
∴f(x)=2√3sin(x+π/3)
Let x + π / 3 = π + K π, K ∈ Z
The center of symmetry is x = 2 π / 3 + K π, K ∈ Z
(2)
f(A)=2√3sin(A+π/3)=3
∴sin(A+π/3)=√3/2
A=π/3
a=BC=√3
Cosine theorem
cosA=(b²+c²-a²)/(2bc)=1/2
∴b²+c²-3=bc
b²+c²=3+bc
∵b²+c²>=2bc
∴3+bc>=2bc
BC

Let f (x) = radical 3cos ^ 2 (Wx) + sinwxcoswx + a (W > 0) and the abscissa of the first highest point of the image of F (x) on the right side of the y-axis is pi / 6

f(x)＝√3/2cos2ωx＋1/2sin2ωx＋√3/2＋a
＝sin(2ωx＋π/3)＋√3/2＋a
According to the meaning of the title:
2ω×(π/6)＋π/3＝π/2
The solution is: ω = 1 / 2
∴f(x)＝sin(x＋π/3)＋√3/2＋a

Let f (x) = sinwxcoswx + root 3cos square Wx (W > 0) have a symmetric center of P (π, root 3 / 2) to find the minimum value of W When the minimum value of W is obtained, the monotone increasing interval of y = Tan (Wx + π / 4) is obtained

The analysis is as follows: ∵ f (x) = sin  s  f (x) = sin  (x) 3 (COS, ω x) ^ 2 = sin2  2 = sin2  3 / 2 = sin (2  2

Given the vector a = (coswx, sinwx), vector b = (coswx, Radix 3coswx), where 0

f(x)=a*b-1/2=(coswx,sinwx)(coswx,√3coswx)-1/2=cos²wx+√3sinwx*coswx-1/2
=1 / 2 * cos2wx + √ 3 / 2sin2wx = sin (2wx + π / 6), and the symmetry axis of the image is x = Pie / 6
If f (0) = f (π / 3), so sin π / 6 = sin (2W π / 3 + π / 6), then
π / 6 = 2W π / 3 + π / 6, or 2W π / 3 + π / 6 = 5 π / 6
W = 0 or w = 1, where 0

M vector = (radical 3sinwx, 0), n vector = (coswx, - sinwx), w > 0, in the image of function f (x) = m (M + n) + T, the minimum distance between the center of symmetry and the axis of symmetry is obtained When x belongs to [0, π / 3], the maximum value of F (x) is 1, (1) find the analytic formula of function f (x), (2) if f (x) = - (1 + radical 3) / 2, X belongs to [0, π], find the value of real number X

(1)f(x)=m(m+n)+t
=m²+mn+t
=3(sinwx)²+√3sinwxcoswx+t
=3/2(1-cos2wx)+(√3/2)sin2wx+t
=√3sin(2wx-π/3)+3/2+t
The minimum distance from the center of symmetry to the axis of symmetry is π / 4
The minimum positive period = 2 π / (2W) = π, w = 1
When x belongs to [0, π / 3], 2x - π / 3 belongs to [- π / 3, π / 3]
If the maximum value of F (x) is 1, then √ 3 × (√ 3 / 2) + 3 / 2 + T = 1
t=-2
∴f(x)=√3sin(2x-π/3)-1/2
(2)f(x)=-（1+√3）/2=√3sin(2x-π/3)-1/2
sin(2x-π/3)=-1/2
Because x belongs to [0, π], x = π / 12,3 π / 4

It is known that the minimum positive period of the function f (x) = (radical 3 * sinwx + coswx) * coswx-1 / 2 (W > 0) is 4 π (1) Find the monotone decreasing interval of F (x); (2) In the triangle ABC, the opposite sides of angles a, B and C are a, B and C respectively, which satisfy (2a-c) CoSb = bcosc. Find the value range of function f (a)

The minimum positive period of the minimum positive period of cos2wx = sin (2wx + 2wx) (W > 0) is 4 π, 2 π / (2W) = 4 π, w = 1 / 4. (1) f (x) = sin (x / 2 + π / 6), w = 1 / 4. (1) f (x) = sin (x / 2 + 2 + π / 6) from (2k + 1 / 2) π (1 / 2) cos2wx = sin (2wx + 2W / 6) (W > 0) (W > 0) is 4 π, 2 π / (2W) = 4 π, w = 1 / 1 / 4. (1) f (x) = sin (x / 2 + 2 + π / 6) has a monotone decreasing interval from (2k + 1 / 2) π π (2k + 1 / 2) π

Given the function f (x) = 2Sin ω x * cos ω x (ω > 0, X ∈ R) (1), find the value range of F (x) (2) if the minimum positive period of F (x) is 4 π,

From the angle doubling formula, f (x) = sin2wx
Then the value range of F (x) is [- 1,1]
T=2π/2w=4π
Results: w = 1 / 4
So, f (x) = sin (x / 2)